A) 1
B) \[-1\]
C) 4
D) no real value
Correct Answer: D
Solution :
Given, \[A=\left[ \begin{matrix} \alpha & 0 \\ 1 & 1 \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 0 \\ 5 & 1 \\ \end{matrix} \right]\]and\[{{A}^{2}}=B\] \[\Rightarrow \] \[{{A}^{2}}=\left[ \begin{matrix} \alpha & 0 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} \alpha & 0 \\ 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} {{\alpha }^{2}} & 0 \\ \alpha +1 & 1 \\ \end{matrix} \right]\] \[\because \] \[{{A}^{2}}=B\] \[\Rightarrow \] \[\left[ \begin{matrix} {{\alpha }^{2}} & 0 \\ \alpha +1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 5 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \] \[{{\alpha }^{2}}=1\]and\[\alpha +1=5\] Which is not possible at the same time. \[\therefore \]No real value of a exist.
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